The equation of a circle $C$ is $x^2+y^2+2x+18y+66 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Solution: To find the equation in standard form, complete the square. $(x^2+2x) + (y^2+18y) = -66$ $(x^2+2x+1) + (y^2+18y+81) = -66 + 1 + 81$ $(x+1)^{2} + (y+9)^{2} = 16 = 4^2$ Thus, $(h, k) = (-1, -9)$ and $r = 4$.